Question

If a 120-W lightbulb emits 2.5 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions.

Part A. How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.9 m away? Answer in photons/s

Part B. How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.5 km away? Answer in photons/s

Answer 1

Given,

P = 120 W ; % = 2.5 ; lambda = 550 nm

A)d = 4 mm ; D = 1.9 m

P = 0.025 x 120 = 3 W

We know that,energy carried by a single photon is:

E = hc/lambda = 6.626 x 10^-34 x 3 x 10^8/(550 x 10^-9) = 3.61 x 10^-19 J

We know that, intensity is:

I = P/A = 3/4 x 3.14 x 1.9^2 = 0.066 W/m^2

P = I x A (at eye)

P = 0.066 x 3.14 x (2 x 10^-3)^2 = 8.3 x 10^-7 J/s

N = 8.3 x 10^-7 J/s/(3.61 x 10^-19) = 2.3 x 10^12 photons/s

Hence, N = 2.3 x 10^12 photons/s

B)I = P/A = 3/4 x 3.14 x 1500^2 = 1.1 x 10^-7 W/m^2

P = I x A (at eye)

P = 1.1 x 10^-7 x 3.14 x (2 x 10^-3)^2 = 1.4 x 10^-12 J/s

N = 1.4 x 10^-12 J/s/(3.61 x 10^-19) = 3.9 x 10^6 photons/s

Hence, N = 3.9 x 10^6 photons/s